Explanation:
You can use Hollomon's equation to estimate this.
Y_f is the flow stress.
K is the strenght coefficient or constant.
e is the strain
n is the strain hardening exponent.
You should be able to solve for e given the data in the problem statement.
Answer:
Explanation:
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Explanation:
There has been no information about related to which programming language is to be used, writing code algorithm.
Defining I/O's ;
Analogue output
A01, A02, A03,AO4
Analogue Input;
AI_1 // potentiometer input
// Based on controller used, assign channels to I/O's
// code
int voltage
Voltage = AI_1;
If (Voltage > 0 && Voltage < 1.25)
{
A01 = voltage
A02 = 0;
A03= 0
AO4= 0
}
If (Voltage > 1.25 && Voltage < 2.5)
{
A01 = 1.25
A02 = (Voltage -1.25);
A03= 0
AO4= 0
}
If (Voltage > 2.5 && Voltage < 3.75)
{
A01 = 1.25
A02 = 1.25
A03= (Voltage - 2.5);
AO4= 0
}
else
{
A01 = 1.25
A02 = 1.25
A03= 1.25
AO4= (Voltage - 3.75);
}
return
you are given the following data set: 15, 12, 11, 13, 10, 23, 16, 18, 11, 8. find the values of the lower quartile, median, uppe
Sav [38]
First quartile: Q1=11
second: Q2=12.5
third: Q3=16
interquartile: 5
median: Q2=12.5
minimum: 8
maximum: 23
range: 15
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = V<em>I</em> = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
